# Exercise II.2.2: Prove that $\lim \dfrac{n^3}{2^n} = 0$ [duplicate]

I'm trying to solve Problem II.2.2 in textbook Analysis I by Amann/Escher.

Could you please verify whether my attempt on (b) and (c) is fine or it contains logical gaps/errors?

My attempt:

From binomial theorem, we have $$2^n = (1+1)^n = \sum_{k=0}^n {n \choose k} = {n \choose 4} + \sum_{k=0 \atop k\neq 4}^n {n \choose k} = n(n-1)(n-2)(n-3) + \sum_{k=0 \atop k\neq 4}^n {n \choose k}$$

As a result, \begin{aligned} \lim \dfrac{n^3}{2^n} &= \lim \dfrac{n^3}{n(n-1)(n-2)(n-3) + \sum_{k=0 \atop k\neq 4}^n {n \choose k}} \\ &= \lim \dfrac{1}{n \left( 1-\dfrac{1}{n} \right) \left(1-\dfrac{2}{n} \right) \left(1-\dfrac{3}{n} \right) + \dfrac{\sum_{k=0 \atop k\neq 4}^n {n \choose k}}{n^3}} \\ &= \lim \dfrac{1}{n + \dfrac{\sum_{k=0 \atop k\neq 4}^n {n \choose k}}{n^3}} = 0\end{aligned}

• Looks fine to me., though your notation is perhaps a bit sloppy. You just wrote $\lim$ where you mean to write $\lim\limits_{n\to\infty}$. As you work with more than one variable or work with real limits instead, that distinction becomes crucial and it is good to be in the habit of labeling it correctly. Commented Aug 6, 2019 at 0:40
• Also, $\binom{n}{4} = \dfrac{n(n-1)(n-2)(n-3)}{\color{red}{24}}$. Although the coefficients don't matter too much here, they still should not have been forgotten. Commented Aug 6, 2019 at 0:42
• Thank you so much @JMoravitz ;) It's my bad. Commented Aug 6, 2019 at 1:06

It is $$2^n=(1+1)^n=\sum_{k=0}^n \binom{n}{k}>\binom{n}{4}$$ for $$n$$ big enough.
$$\binom{n}{4}=\frac{n!}{4!(n-4)!}=\frac{n(n-1)(n-2)(n-3)}{24}<\frac{n^4}{24}$$
$$\lim_{n\to\infty} \frac{n^3}{2^n}\leq\lim_{n\to\infty} \frac{n^3}{n^4/24}=\lim_{n\to\infty} \frac{24}{n}=0$$
Because you have certainly already proven, that $$\lim_{n\to\infty} \frac1n=0$$